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Math Help - Linear equation problem

  1. #1
    Newbie
    Joined
    Aug 2008
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    Linear equation problem

    Hello everyone. I just don't get this . This seems very basic but I have problems to find a answer to this. I know augmented matrices and gaussian elimination.

    Ok once again Ax = b
    Okay I get the augmented matrix form. But how can I do this "backwards"

    A is 2x2 matrix , and
    x =[0;
    3]
    b = Null vector

    if A was [a,b;c,d] then I came up with 3c + 3d = 0 and then set c =1 ,d=-1
    This didn't function

    At the moment I am incapable of coming up with a matrix which certain vectors are in the kernel or image. I can't make a matrix which needs to have certain solutions.
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  2. #2
    Member
    Joined
    Aug 2008
    Posts
    80
    Hello,

    If you mean:
    Ax=b where <br />
A=\begin{pmatrix}a&b\\c&d\end{pmatrix},\, <br />
x=\begin{pmatrix}0\\3\end{pmatrix},\, <br />
b=\begin{pmatrix}0\\0\end{pmatrix},
    you come up with:
    3b=0,\, 3d=0.
    Thus, A=\begin{pmatrix}a&0\\c&0\end{pmatrix} where a,\, c are arbitrary.

    Bye.
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  3. #3
    Newbie
    Joined
    Aug 2008
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    Exactly what I meant.
    3b=0,\, 3d=0. You got this by the multiplication A=\begin{pmatrix}a&0\\c&0\end{pmatrix} * x=\begin{pmatrix}0\\3\end{pmatrix} which should result in the 2x1 matrix
    3b=0,\, 3d=0.

    To solve those backwards , I take a*b and then rref it? Ok and that would then be your augmented matrix where for example 3b= say 4 if that was what I'd want

    Ok I think I got this, thank you.
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